If (2a+b)(a+4b)=3\frac{(2a+b)}{(a+4b)}=3(a+4b)(2a+b)=3, then find the value of a+ba+2b\frac{a+b}{a+2b}a+2ba+b A27\frac{2}{7}72B\fra{9}{7}C53\frac{5}{3}35D\fra{10}{9}Answer: \fra{10}{9} Read Explanation: Solution:Given 2a+ba+4b=3\frac{2a+b}{a+4b}=3a+4b2a+b=32a+b=3×(a+4b)2a+b=3\times{(a+4b)}2a+b=3×(a+4b)2a+b=3a+12b2a+b=3a+12b2a+b=3a+12ba=−11ba=-11ba=−11bSubstituting this in a+ba+2b\frac{a+b}{a+2b}a+2ba+b=−11b+b−11b+2b=\frac{-11b+b}{-11b+2b}=−11b+2b−11b+b=−10b−9b=\frac{-10b}{-9b}=−9b−10b=109=\frac{10}{9}=910Hence Option(D) is the correct answer. Read more in App